PX262 - F2a - particle in an infinite potential well

s- consider a potential well in 3D:

V (\vec{r}) = {\begin{cases} 0 & i f & - a \leq x \leq a, - b \leq y \leq b and - c \leq z \leq c, \\ \infty & elsewhere . \end{cases}

using separation of variables for the wavefunction: $ϕ (\vec{r}) = X (x) Y (y) Z (z)$
substituting into the time-independent schrödinger equation:

\begin{aligned} \frac{- ℏ^{2}}{2 m} [Y Z \frac{\partial^{2} X}{\partial x^{2}} + X Z \frac{\partial^{2} Y}{\partial y^{2}} + X Y \frac{\partial^{2} Z}{\partial z^{2}}] & = E X Y Z \\ \underset{only dependent on x}{\underset{⏟}{\frac{- ℏ^{2}}{2 m} \frac{1}{X} \frac{\partial^{2} X}{\partial x^{2}}}} + \underset{only dependent on y}{\underset{⏟}{\frac{- ℏ^{2}}{2 m} \frac{1}{Y} \frac{\partial^{2} Y}{\partial y^{2}}}} + \underset{only dependent on z}{\underset{⏟}{\frac{- ℏ^{2}}{2 m} \frac{1}{Z} \frac{\partial^{2} Z}{\partial z^{2}}}} & = E \end{aligned}

each term should be equal to a constant, and the three constants should add up to $E$
looking at the $x$ -dependent term: $- \frac{ℏ^{2}}{2 m} \frac{\partial^{2} X}{\partial x^{2}} = E_{1} X$
this is exactly the same as the equation for the 1D potential well, and the solution is: $E_{n_{1}} = \frac{ℏ^{2} π^{2}}{8 M a^{2}} n_{1}^{2}$ where, $n_{1}$ is the quantum number labelling states
and the wavefunction is: $X (x) = \frac{1}{\sqrt{a}} {\begin{matrix} \cos \\ \sin \end{matrix}} (\frac{n_{1} π x}{2 a})$ where, ${\begin{matrix} \cos \\ \sin \end{matrix}}$ means cosine for odd and sine for even values of $n_{1}$
the full solution is: $ϕ (\vec{r}) = \frac{1}{\sqrt{a b c}} {\begin{matrix} \cos \\ \sin \end{matrix}} (\frac{n_{1} π x}{2 a}) {\begin{matrix} \cos \\ \sin \end{matrix}} (\frac{n_{2} π y}{2 a}) {\begin{matrix} \cos \\ \sin \end{matrix}} (\frac{n_{3} π z}{2 a})$
the energy eigenvalues are the sums of the three contributions: $E = E_{n_{1}} + E_{n_{2}} + E_{n_{3}} = \frac{ℏ^{2} π^{2}}{8 M} (\frac{n_{1}^{2}}{a^{2}} + \frac{n_{2}^{2}}{b^{2}} + \frac{n_{3}^{2}}{c^{2}})$